Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

May $Z$ be a random variable distributed as $N(0,1)$ Find the following limit: \begin{equation} \lim_{n\to\infty}\mathbb{E}\left[\frac{1}{\sqrt{n}-Z}\right] \end{equation}

How does one go about proving it?

share|cite|improve this question
up vote 1 down vote accepted

The expected value does not exist: the function $ \dfrac{f(z)}{\sqrt{n}-z}$ (where $f$ is the probability density function) is not absolutely integrable because of the singularity at $z=\sqrt{n}$. However, the Cauchy principal value of this improper integral does exist. That is, if $I_r(t) = 1$ for $|t|\ge r$ and $0$ for $|t|<r$, ${\mathbb E}\left[ \dfrac{I_r(\sqrt{n}-Z)}{\sqrt{n}-Z} \right]$ exists, and for the limit of this as $r \to 0$ I get $\sqrt{\dfrac{\pi}{2}} e^{-n/2} \text{erfi}(\sqrt{n/2})$, which goes to $0$ as $n \to \infty$.

share|cite|improve this answer
Thanks, V.P. of the integral is what one wants in this case. – Alex Lomachenko Dec 13 '12 at 13:41

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.